What is the integral of x/(sqrt(9+4x)) ?

The integral of x/(sqrt(9+4x)) is 1/12 (2x-9)sqrt(4x+9)+C

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integral of x/(sqrt(9+4x))

\int \frac{x}{9 + 4 x} d x

\frac{1}{12} (2 x - 9) 4 x + 9 + C

Solution steps

\int \frac{x}{9 + 4 x} d x

Apply u-substitution

= \int \frac{u ^{2} - 9}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int u^{2} - 9 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int u^{2} d u - \int 9 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 9 d u = 9 u

= \frac{1}{8} (\frac{u ^{3}}{3} - 9 u)

Substitute back

u = 9 + 4 x

= \frac{1}{8} (\frac{( 9 + 4 x ) ^{3}}{3} - 9 9 + 4 x)

Simplify

\frac{1}{8} (\frac{( 9 + 4 x ) ^{3}}{3} - 9 9 + 4 x) : \frac{1}{12} (2 x - 9) 4 x + 9

= \frac{1}{12} (2 x - 9) 4 x + 9

Add a constant to the solution

= \frac{1}{12} (2 x - 9) 4 x + 9 + C

What is the integral of x/(sqrt(9+4x)) ?
The integral of x/(sqrt(9+4x)) is 1/12 (2x-9)sqrt(4x+9)+C